Page No 59:

Question 6:

A number
is divisible by both 5 and 12. By which other number will that number
be always divisible?

Answer:

Factors of 5 = 1, 5

Factors of 12 = 1, 2, 3, 4, 6, 12

As the
common factor of these numbers is 1, the given two numbers are co-
prime and the number will also be divisible by their product, i.e.
60, and the factors of 60, i.e., 1, 2, 3, 4, 5, 6, 10, 12, 15, 20,
30, 60.

Page No 59:

Question 7:

A number is divisible by 12. By what other number will that number be
divisible?

Answer:

Since the
number is divisible by 12, it will also be divisible by its factors
i.e., 1, 2, 3, 4, 6, 12. Clearly, 1, 2, 3, 4, and 6 are numbers other
than 12 by which this number is also divisible.

Page No 61:

Question 1:

Which of
the following statements are true?

(a) If a
number is divisible by 3, it must be divisible by 9.

(b) If a
number is divisible by 9, it must be divisible by 3.

(c) A
number is divisible by 18, if it is divisible by both 3 and 6.

(d) If a
number is divisible by 9 and 10 both, then it must be divisible by
90.

(e) If two
numbers are co-primes, at least one of them must be prime.

(f) All
numbers which are divisible by 4 must also be divisible by 8.

(g) All
numbers which are divisible by 8 must also be divisible by 4.

(h) If a number exactly divides two numbers separately, it must
exactly divide their sum.

(i) If a number exactly divides the sum of two numbers, it must
exactly divide the two numbers separately.

Answer:

(a) False

6 is divisible by 3, but not by 9.

(b) True,
as 9 = 3 × 3

Therefore, if a number is divisible by 9, then it will also be
divisible by

3.

(c) False

30 is divisible by 3 and 6 both, but it is not divisible by 18.

(d) True,
as 9 × 10 = 90

Therefore, if a number is divisible by 9 and 10 both, then it will
also be divisible by 90.

(e) False

15 and 32 are co-primes and also composite.

(f) False

12 is divisible by 4, but not by 8.

(g) True,
as 8 = 2 × 4

Therefore, if a number is divisible by 8, then it will also be
divisible by 2 and 4.

(h) True

2 divides 4 and 8 as well as 12. (4 + 8 = 12)

(i) False

2 divides 12, but does not divide 7 and 5.

Page No 62:

Question 2:

Here are
two different factor trees for 60. Write the missing numbers.

(a)

(b)

Answer:

(a) As 6 =
2 × 3 and 10 = 5 × 2

(b) As 60
= 30 × 2, 30 = 10 × 3, and 10 = 5 × 2

Page No 62:

Question 3:

Which factors are not included in the prime factorization of a
composite number?

Answer:

1 and the
number itself

Page No 62:

Question 4:

Write the
greatest 4-digit number and express it in terms of its prime factors.

Answer:

Greatest
four-digit number = 9999

9999 = 3 ×
3 × 11 × 101

Page No 62:

Question 5:

Write the smallest 5-digit number and express it in the form of its
prime factors.

Answer:

Smallest
five-digit number = 10,000

10000 = 2
× 2 × 2 × 2 × 5 × 5 × 5 × 5

Page No 62:

Question 6:

Find all prime factors of 1729 and arrange them in ascending
order. Now state the relation, if any; between two consecutive prime
factors.

Answer:

7

1729

13

247

19

19

1

1729 = 7 × 13 × 19

13 − 7 = 6, 19 − 13 = 6

The difference of two consecutive prime factors is 6.

Page No 62:

Question 7:

The
product of three consecutive numbers is always divisible by 6. Verify
this statement with the help of some examples.

Answer:

2 ×
3 × 4 = 24, which is divisible by 6

9 ×
10 × 11 = 990, which is divisible by 6

20 ×
21 × 22 = 9240, which is divisible by 6

Page No 62:

Question 8:

The sum of
two consecutive odd numbers is divisible by 4. Verify this statement
with the help of some examples.

Answer:

3 + 5 = 8,
which is divisible by 4

15 + 17 =
32, which is divisible by 4

19 + 21 =
40, which is divisible by 4

Page No 62:

Question 9:

In which of the following expressions, prime factorization has been
done?

(a) 24 = 2 × 3 ×
4 (b) 56 = 7 × 2 ×
2 × 2

(c) 70 = 2
× 5 ×
7 (d) 54 = 2 × 3 ×
9

Answer:

(a) 24 = 2
× 3 × 4

Since 4 is composite, prime factorisation has not been done.

(b) 56 = 7
× 2 × 2 × 2

Since all the factors are prime, prime factorisation has been done.

(c) 70 = 2
× 5 × 7

Since all the factors are prime, prime factorisation has been done.

(d) 54 = 2
× 3 × 9

Since 9 is composite, prime factorisation has not been done.

Page No 62:

Question 10:

Determine if 25110 is divisible by 45.

[Hint: 5
and 9 are co-prime numbers. Test the divisibility of the number by 5
and 9].

Answer:

45 = 5 ×
9

Factors of
5 = 1, 5

Factors of
9 = 1, 3, 9

Therefore,
5 and 9 are co-prime numbers.

Since the
last digit of 25110 is 0, it is divisible by 5.

Sum of the
digits of 25110 = 2 + 5 + 1 + 1 + 0 = 9

As the sum
of the digits of 25110 is divisible by 9, therefore, 25110 is
divisible by 9.

Since the
number is divisible by 5 and 9 both, it is divisible by 45.

Page No 62:

Question 11:

18 is
divisible by both 2 and 3. It is also divisible by 2 ×
3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say
that the number must also be divisible by 4 ×
6 = 24? If not, give an example to justify our answer:

Answer:

No. It is
not necessary because 12 and 36 are divisible by 4 and 6 both, but
are not divisible by 24.

Page No 62:

Question 12:

I am the smallest number, having four different prime factors. Can
you find me?

Answer:

Since it
is the smallest number of such type, it will be the product of 4
smallest prime numbers.

2 ×
3 × 5 × 7 = 210

Page No 63:

Question 1:

Find the
HCF of the following numbers:

(a) 18,
48 (b) 30, 42 (c) 18, 60

(d) 27,
63 (e) 36, 84 (f) 34, 102

(g) 70,
105, 175 (h) 91, 112, 49 (i) 18, 54, 81

(j) 12,
45, 75

Answer:

(a) 18, 48

2

18

3

9

3

3

1

2

48

2

24

2

12

2

6

3

3

1

18 = 2 × 3 × 3

48 = 2 × 2 × 2 × 2 × 3

HCF = 2 × 3 = 6

(b) 30, 42

2

30

3

15

5

5

1

2

42

3

21

7

7

1

30 = 2 × 3 × 5

42 = 2 × 3 × 7

HCF = 2 × 3 = 6

(c) 18, 60

2

18

3

9

3

3

1

2

60

2

30

3

15

5

5

1

18 = 2 × 3 × 3

60 = 2 × 2 × 3 × 5

HCF = 2 × 3 = 6

(d) 27, 63

3

27

3

9

3

3

1

3

63

3

21

7

7

1

27 = 3 × 3 × 3

63 = 3 × 3 × 7

HCF = 3 × 3 = 9

(e) 36, 84

2

36

2

18

3

9

3

3

1

2

84

2

42

3

21

7

7

1

36 = 2 × 2 × 3 × 3

84 = 2 × 2 × 3 × 7

HCF = 2 × 2 × 3 = 12

(f) 34,
102

2

34

17

17

1

2

102

3

51

17

17

1

34 = 2 × 17

102 = 2 × 3 × 17

HCF = 2 ×17 = 34

(g) 70,
105, 175

2

70

5

35

7

7

1

3

105

5

35

7

7

1

5

175

5

35

7

7

1

70 = 2 × 5 × 7

105 = 3 × 5 × 7

175 = 5 × 5 × 7

HCF = 5 × 7 = 35

(h) 91,
112, 49

7

91

13

13

1

2

112

2

56

2

28

2

14

7

7

1

7

49

7

7

1

91 = 7 × 13

112 = 2 × 2 × 2 × 2 × 7

49 = 7 × 7

HCF = 7

(i) 18,
54, 81

2

18

3

9

3

3

1

2

54

3

27

3

9

3

3

1

3

81

3

27

3

9

3

3

1

18 = 2 × 3 × 3

54 = 2 × 3 × 3 × 3

81 = 3 × 3 × 3 × 3

HCF = 3 × 3 = 9

(j) 12,
45, 75

2

12

2

6

3

3

1

3

45

3

15

5

5

1

3

75

5

25

5

5

1

12 = 2 ×2 × 3

45 = 3 × 3 × 5

75 = 3 × 5 × 5

HCF = 3

Page No 63:

Question 2:

What is the HCF of two consecutive

(a) Numbers? (b) Even numbers? (c) Odd numbers?

Answer:

(i) 1 e.g., HCF of 2 and 3 is 1.

(ii) 2 e.g., HCF of 2 and 4 is 2.

(iii) 1 e.g., HCF of 3 and 5 is 1.

Page No 64:

Question 3:

HCF of
co-prime numbers 4 and 15 was found as follows by factorization:

4 = 2 ×
2 and 15 = 3 × 5 since there
is no common prime factors, so HCF of 4 and 15 is 0. Is the answer
correct? If not, what is the correct HCF?

Answer:

No. The
answer is not correct. 1 is the correct HCF.

Page No 67:

Question 1:

Renu
purchases two bags of fertilizer of weight 75 kg and 69 kg. Find the
maximum value of weight which can measure the weight of the
fertilizer exact number of times.

Answer:

Weight of
the two bags = 75 kg and 69 kg

Maximum
weight = HCF (75, 69)

3

75

5

25

5

5

1

3

69

23

23

1

75 = 3 ×
5 × 5

69 = 3 ×
23

HCF = 3

Hence, the
maximum value of weight, which can measure the weight of the
fertilizer exact number of times, is 3 kg.

Page No 67:

Question 2:

Three boys
step off together from the same spot. Their steps measure 63 cm, 70
cm and 77 cm respectively. What is the minimum distance each should
cover so that all can cover the distance in complete steps?

Answer:

Step
measure of 1st Boy = 63 cm

Step
measure of 2nd Boy = 70 cm

Step
measure of 3rd Boy = 77 cm

LCM of 63,
70, 77

2

63, 70, 77

3

63, 35, 77

3

21, 35, 77

5

7, 35, 77

7

7, 7, 77

11

1, 1, 11

1, 1, 1

LCM = 2 ×
3 × 3 × 5 × 7 × 11 = 6930

Hence, the
minimum distance each should cover so that all can cover the distance
in complete steps is 6930 cm.

Page No 67:

Question 3:

The
length, breadth and height of a room are 825 cm, 675 cm and 450 cm
respectively. Find the longest tape which can measure the three
dimensions of the room exactly.

Answer:

Length =
825 cm = 3 × 5 × 5 × 11

Breadth =
675 cm = 3 × 3 × 3 × 5 × 5

Height =
450 cm = 2 × 3 × 3 × 5 × 5

Longest
tape = HCF of 825, 675, and 450 = 3 × 5 × 5 = 75 cm

Therefore,
the longest tape is 75 cm.

Page No 67:

Question 4:

Determine the smallest 3-digit number which is exactly divisible by
6, 8 and 12.

Answer:

Smallest
number = LCM of 6, 8, 12

2

6, 8, 12

2

3, 4, 6

2

3, 2, 3

3

3, 1, 3

1, 1, 1

LCM = 2 ×
2 × 2 × 3 = 24

We have to
find the smallest 3-digit multiple of 24.

It can be
seen that 24 × 4 = 96 and 24 × 5 = 120.

Hence, the
smallest 3-digit number which is exactly divisible by 6, 8, and 12 is
120.

Page No 67:

Question 5:

Determine the greatest 3-digit number exactly divisible by 8, 10 and
12.

Answer:

LCM of 8,
10, and 12

2

8, 10, 12

2

4, 5, 6

2

2, 5, 3

3

1, 5, 3

5

1, 5, 1

1,1,1

LCM = 2 ×
2 × 2 × 3 × 5 = 120

We have to
find the greatest 3-digit multiple of 120.

It can be
seen that 120 ×8 = 960 and 120 × 9 = 1080.

Hence, the greatest 3-digit number exactly divisible by 8, 10, and 12
is 960.

Page No 67:

Question 6:

The
traffic lights at three different road crossings change after every
48 seconds, 72 seconds and 108 seconds respectively. If they change
simultaneously at 7 a.m., at what time will they change
simultaneously again?

Answer:

Time
period after which these lights will change = LCM of 48, 72, 108

2

48, 72, 108

2

24, 36, 54

2

12, 18, 27

2

6, 9, 27

3

3, 9, 27

3

1, 3, 9

3

1, 1, 3

1, 1, 1

LCM = 2 ×
2 × 2 × 2 × 3 × 3 × 3 = 432

They will
change together after every 432 seconds i.e., 7 min 12 seconds.

Hence,
they will change simultaneously at 7:07:12 am.

Page No 67:

Question 7:

Three
tankers contain 403 litres, 434 litres and 465 litres of diesel
respectively. Find the maximum capacity of a container that can
measure the diesel of the three containers exact number of times.

Answer:

Maximum
capacity of the required tanker = HCF of 403, 434, 465

403 = 13 ×
31

434 = 2 ×
7 × 31

465 = 3 ×
5 × 31

HCF = 31

∴A
container of capacity 31 l can measure the diesel of 3
containers exact number of times

Page No 67:

Question 8:

Find the
least number which when divided by 6, 15 and 18 leave remainder 5 in
each case.

Answer:

LCM of 6,
15, 18

2

6, 15, 18

3

3, 15, 9

3

1, 5, 3

5

1, 5, 1

1, 1, 1

LCM = 2 ×
3 × 3 × 5 = 90

Required
number = 90 + 5 = 95

Page No 67:

Question 9:

Find the smallest 4-digit number which is divisible by 18, 24 and 32.

Answer:

LCM of 18,
24, and 32

2

18, 24, 32

2

9, 12, 16

2

9, 6, 8

2

9, 3, 4

2

9, 3, 2

3

9, 3, 1

3

3, 1, 1

1, 1, 1

LCM = 2 ×
2 × 2 × 2 × 2 × 3 × 3 = 288

We have to
find the smallest 4-digit multiple of 288.

It can be
observed that 288 ×3 = 864 and 288 ×4 = 1152.

Therefore,
the smallest 4-digit number which is divisible by 18, 24, and 32 is

1152.

Page No 67:

Question 10:

Find the LCM of the following numbers:

(a) 9 and 4 (b) 12 and 5

(c) 6 and
5 (d) 15 and 4

Observe a
common property in the obtained LCMs. Is LCM the product of two
numbers in each case?

Answer:

(a)

2

9, 4

2

9, 2

3

9, 1

3

3, 1

1, 1

LCM = 2 × 2 × 3 × 3 = 36

(b)

2

12, 5

2

6, 5

3

3, 5

5

1, 5

1, 1

LCM = 2 ×
2 × 3 × 5 = 60

(c)

2

6, 5

3

3, 5

5

1, 5

1, 1

LCM = 2 ×
3 × 5 = 30

(d)

2

15, 4

2

15, 2

3

15, 1

5

5, 1

1, 1

LCM = 2 × 2 × 3 × 5 = 60

Yes, it can be observed that in each case, the LCM of the given
numbers is the product of these numbers. When two numbers are
co-prime, their LCM is the product of those numbers. Also, in each
case, LCM is a multiple of 3.

Page No 67:

Question 11:

Find the
LCM of the following numbers in which one number is the factor of the
other.

(a) 5, 20 (b) 6, 18

(c) 12,
48 (d) 9, 45

What do
you observe in the results obtained?

Answer:

(a)

2

5, 20

2

5, 10

5

5, 5

1, 1

LCM = 2 ×
2 × 5 = 20

(b)

2

6, 18

3

3, 9

3

1, 3

1, 1

LCM = 2 ×
3 × 3 = 18

(c)

2

12, 48

2

6, 24

2

3, 12

2

3, 6

3

3, 3

1, 1

LCM = 2 ×
2 × 2 × 2 × 3 = 48

(d)

3

9, 45

3

3, 15

5

1, 5

1, 1

LCM = 3 ×
3 × 5 = 45

Yes, it
can be observed that in each case, the LCM of the given numbers is
the larger number. When one number is a factor of the other number,
their LCM will be the larger number.